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Confused by Pair Production

  1. Sep 28, 2013 #1
    1. The problem statement, all variables and given/known data

    How much photon energy would be required to produce a proton-antiproton pair? Where could such a
    high-energy photon come from?

    2. Relevant equations

    [itex]E^{2} = (pc)^{2} + (E_{0})^{2}[/itex]

    3. The attempt at a solution

    Since the photon is massless its energy is
    [itex]E_{i}=pc[/itex]

    The final energy is:
    [itex]E_{f}^{2}=(pc)^{2}+(2m_{+}c^{2})^2[/itex]

    and we know that energy and momentum are conserved, so:
    [itex]E_{i}^{2}=E_{f}^{2}[/itex]
    [itex](pc)^{2} = (pc)^{2}+(2m_{+}c^{2})^2[/itex]

    but clearly, the mass of a photon is not zero, so how can this be?

    edit-
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    Please.. Can someone fix this? This gets added at the end of my post every time I preview post. It has been doing this for years...
     
  2. jcsd
  3. Sep 29, 2013 #2

    UltrafastPED

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    You need at least twice the mass of the proton, and this must be delivered within a small area and a small time.

    For example, we can easily make electron-positron pairs with an ultrafast laser (20 fs pulses) whose average power is less than a watt, and generates a thousand pulses per second. Thus each pulse has a milli-Joule of energy, but by using a focus-correcting scheme it can all be put into a tiny area in a very brief time. Smash this into a gas or a solid target to get the energy out of the light pulse, and you will find electron positron pairs.

    If you use a bigger laser, you can get proton-anti-proton pairs.

    See http://cuos.engin.umich.edu/researchgroups/hfs/facilities/hercules-petawatt-laser/
     
  4. Sep 29, 2013 #3
    Oh ok, this question is asking about a single photon though, not a thousand photons per second.

    So you're saying that in order for pair production to happen, the photon must collide into something and lose its momentum? Is it fair to say that momentum can be completely lost during the collision and two particles at rest can be created? Would this be the mimimum photon energy necessary for pair production to happen? If so, I think that answers my question and I can solve this problem.
     
  5. Sep 29, 2013 #4

    UltrafastPED

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    If your particle has a rest mass of 2 GeV/c^2 you will need a photon of double that energy.

    If you don't cancel all of the momentum then you will need more to offset it ... photons also have momentum.

    Thus it is better to hit a solid target, or else you need to bring two photons head-to-head, but you also need a moderator because photons don't interact at that energy level.

    And conceptually there is no difference between one giant photon and a million small ones seen at the same time ... they both deliver the energy at the same rate, and in the same volume. Thus if I send a bunch of 1.5 eV photons (800 nm, near infrared) at a metal film with a 4.5 eV work function ... why I will get multi-photon absorption, and yes, I will get photo-electrons.

    This is well documented, and I have also done it myself. My lab mates have created the electron-positron pairs, and the lab across the hall (Hercules) has generated proton beams, transmuted elements, etc ... all with a table-top laser ... its a big table, but it does fit in one room.
     
  6. Sep 29, 2013 #5

    vela

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    Where ##p## is the momentum of the photon.

    What does ##p## represent here? It's not the momentum of the photon.


    An isolated photon cannot pair produce because you can't conserve both energy and momentum. Think about the reverse process, proton-anti proton annihilation. In the center of mass frame, the total momentum initially is zero, but if the annihilation produced only one photon, there would be a non-zero total momentum afterward. There has to be another photon produced to conserve momentum.

    With pair production, the second photon is typically a virtual photon exchanged with a nearby nucleus. The nucleus ends up absorbing the extra momentum without taking much energy.
     
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